A /\ /  \ / e \ /  \ h  i / D \ / f g \ /   \ BjC 
Place ten numbers at the four corners and six connecting edges of a tetrahedron, such that either:

Source: Original.
For the edges, let S = the common sum. Then the sum of all 6
lines = 6S = the sum of all the numbers + twice the sum of the corners.
For minimum sum, we want to put the smallest numbers on the corners. But we
cannot put 4 consecutive numbers on the corners, because then A+D = B+C and
every pair of corners has an edge, so the middle numbers on those edges would be
forced to be the same. If we put 1,2,3,5 on the corners, then the pairsums of
the corners are 3, 4, 5, 6, 7, and 8, and the edges can be the first 6
consecutive unused numbers: 6 to 11. Common sum = 14.
So A,B,C,D = 5,3,2,1, and then e,f,g,h,i,j = 8,10,11,6,7,9.
Any other set of numbers for the corners adds to at least 12, and with the sum
of all the numbers at a minimum (for 1 through 10) sum of 55, 6S would be at
least 79, which is more than 6 times 13, so 14 is the smallest possible common
sum. Since the sum of all the numbers 1 to 10 is odd, and 6S = this odd sum plus
2 times something, it is not possible to make a sum of 14 with the numbers 1 to
10, so the above case is the minimal solution.
For the faces, let T = the common sum. Then the sum of all faces = 4T = twice
the sum of all numbers + the sum of the corners. We still want to put the small
numbers on the corners. Since there are three numbers getting added to each trio
of corners, it's not immediately clear whether simply letting the corners be 1,
2, 3, and 4 will work, but if so, we will need edges which combine in various
ways to sums that are 4 consecutive numbers.
If the edges are 5 through 10, then T=30, and we need the edges on the faces to
add to 21, 22, 23, and 24. If 10 and 9 are on the same face, then only 10+9+5=24
and 10+7+6=23 will work. Otherwise, 9 is opposite 10, and 10 must form two faces
with the other numbers, which can only be 10+8+6=24 on one face and 10+7+5=22 on
the other. In each case, the missing number must join with the four numbers
other than 10 to form the other two faces, using one number from each of the
first two faces on each of the remaining two. This means that one of these faces
will contain both 8 and 9, and in each case there is no third number available
which will form a sum not already used.
So the sum must be larger than 30. The next smallest sum possible is 31, and it
makes sense to spread out the numbers a bit, so try it using numbers 1 to 10,
but corners which add to a sum of 14.
There are several possibilities, and I first tried 1, 3, 4, 6 and 1, 2, 5, 6
which both failed. Try 1, 2, 4, 7 for the corners. Now the middle numbers are 3,
5, 6, 8, 9, 10, and the sums of these on the faces are 18, 19, 21, and 24. This
works with 10+9+5=24 and 10+8+3=21 on two faces, 6+9+3=18 on one face and
6+8+5=19 on the last.
So we have A,B,C,D = 7,4,2,1. The edge assignments work out to e,f,g,h,i,j =
8,5,10,6,3,9.
A = 5, B = 3, C = 2, D = 1 e = 8, f = 10, g = 11, h = 6, i = 7, j = 9 5 /\ /  \ / 8 \ /  \ 6  7 / 1 \ / 10 11 \ /   \ 392
Part 1: max number = 10 max sum = 11 0 /\ /  \ / 7 \ /  \ 10  9 / 4 \ / 6 5 \ /   \ 182 11 = Sum 0 1 2 4 7 6 5 10 9 8 0 /\ /  \ / 5 \ /  \ 10  9 / 6 \ / 4 3 \ /   \ 182 11 = Sum 0 1 2 6 5 4 3 10 9 8 0 /\ /  \ / 5 \ /  \ 10  8 / 6 \ / 4 2 \ /   \ 173 11 = Sum 0 1 3 6 5 4 2 10 8 7 Part 2: max number = 9 max sum = 23 0 /\ /  \ / 5 \ /  \ 8  9 / 3 \ / 6 4 \ /   \ 172 23 = Sum 0 1 2 3 5 6 4 8 9 7 0 /\ /  \ / 6 \ /  \ 9  7 / 3 \ / 4 5 \ /   \ 182 23 = Sum 0 1 2 3 6 4 5 9 7 8